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alakaboom1
02-14-2011, 02:28 PM
An automobile insurance company classifies drivers into 3 classes: class A, class B, and class C. The percentage of drivers in each class is: class A, 20%; class B, 65%; and class C, 15%. The probabilities that a driver in one of these classes will have an accident within one year are given by 0.01, 0.02, and 0.03 respectively. After purchasing an insurance policy, a driver has an accident within the first year. What is the probability that the driver is of class A?

I think that it's 1/6, because you'd do 0.01/0.06, since you already know that the driver has had the accident, but I'm not too sure. One other possibility that I'm considering would be 0.2 * 0.01, but I feel like this doesn't account for the fact that we are assuming that the driver is already in an accident...

galactus
02-14-2011, 03:23 PM
You are looking for the probability the driver is from class A given they have an accident:

P(\text{class A}|\text{accident})

You can make a chart. Assume a number of drivers that is easy to work with. Say, 1000

Or, use Bayes Theorem. Do you know it?.

soroban
02-14-2011, 08:19 PM
Hello, alakaboom1!

Here is a very primitive approach . . .

An automobile insurance company classifies drivers into 3 classes: class A, class B, and class C.
The percentage of drivers in each class is: class A, 20%; class B, 65%; and class C, 15%.
The probabilities that a driver in one of these classes will have an accident within one year are given by 0.01, 0.02, and 0.03 respectively.
After purchasing an insurance policy, a driver has an accident within the first year.
What is the probability that the driver is of class A?
Suppose there are 10,000 policyholders.

\text{20\% are class A drivers: }\:20\%\times 10,\!000 \,=\,2000
. . \text{1\% of them will have an accident: }\:1\% \times 2000 \,=\,20

\text{65\% are class B drivers: }\:65\% \times 10,\!000 \,=\,6500
. . \text{2\% of them will have an accident: }\:2\% \times 6500 \,=\,130

\text{15\% are class C drivers: }\:15\% \times 10,\!000 \,=\,1500
. . \text{3\% of them will have an accident: }\:3\% \times 1500 \,=\,45

\text{There is a total of: }\:20 + 130 + 45 \,=\,195\text{ accidents}
. . \text{of which 20 are class A drivers.}

\text{Therefore: }\:P(\text{class A}\,|\,\text{accident}) \;=\;\frac{20}{195} \;=\;\frac{4}{39}

galactus
02-14-2011, 09:11 PM
That 'primitive
approach is a nice explanation.

Here is all I was getting at:

\frac{(.01)(.2)}{(.01)(.2)+(.02)(.65)+(.03)(.15)}= \frac{4}{39}\approx .1025

When confronted with a 'probability of something given something' problem, you can make a chart.

Assume 1000 policyholders.

\begin{array}{c|c|c|c|c}\text{}&A&B&C&\text{total}\\ \hline\text{has accidents}&2&13&4.5&19.5 \\ \hline\text{no accident}&198&637&145.5&980.5 \\ \hline\text{total}&200&650&150&1000\end{array}

Now, you can answer any problem they throw at you. In this case, we want

"Probability the driver is class A given they have an accident".

Go down the class A column and across the 'has accident' row. \frac{2}{19.5}=\frac{4}{39}

Say they asked for "probability the driver is from class C given they did not have an accident?"

Go down the C column and across the 'no accident' row and get \frac{145.5}{980.5}=\frac{291}{1961}\approx .148

and so on..............

alakaboom1
02-14-2011, 10:18 PM
Thanks a lot guys!

Out of curiosity, is this formula you guys mentioned known as Bayes Theorem? I've definitely used the formula before, but never knew it by name.